Q123 / 150 · MHT CET 2026 (17 April Shift 1) · Physics
For a particle executing S.H.M., its potential energy is 8 times its kinetic energy at a certain displacement '$x$' from the mean position. If '$A$' is the amplitude of S.H.M., the value of '$x$' is
- A$\dfrac{2}{\sqrt{3}}A$
- B$\dfrac{\sqrt{2}}{3}A$
- Correct answerC$\dfrac{2\sqrt{2}}{3}A$
- D$\dfrac{3}{\sqrt{2}}A$
Solution
The potential energy of a particle executing S.H.M. at a displacement $x$ from the mean position is given by $U = \dfrac{1}{2}m\omega^2x^2$
The kinetic energy of the particle at displacement $x$ is given by $K = \dfrac{1}{2}m\omega^2(A^2 - x^2)$
Given that the potential energy is 8 times the kinetic energy:
$U = 8K$
$\dfrac{1}{2}m\omega^2x^2 = 8 \times \dfrac{1}{2}m\omega^2(A^2 - x^2)$
$x^2 = 8(A^2 - x^2)$
$x^2 = 8A^2 - 8x^2$
$9x^2 = 8A^2$
$x^2 = \dfrac{8}{9}A^2$
$x = \dfrac{2\sqrt{2}}{3}A$
Answer: $\dfrac{2\sqrt{2}}{3}A$
The kinetic energy of the particle at displacement $x$ is given by $K = \dfrac{1}{2}m\omega^2(A^2 - x^2)$
Given that the potential energy is 8 times the kinetic energy:
$U = 8K$
$\dfrac{1}{2}m\omega^2x^2 = 8 \times \dfrac{1}{2}m\omega^2(A^2 - x^2)$
$x^2 = 8(A^2 - x^2)$
$x^2 = 8A^2 - 8x^2$
$9x^2 = 8A^2$
$x^2 = \dfrac{8}{9}A^2$
$x = \dfrac{2\sqrt{2}}{3}A$
Answer: $\dfrac{2\sqrt{2}}{3}A$
Hence, the correct answer is option: 'C'.